Find $\displaystyle \int_0^1\frac{x^2}{(1-x)\ln^2(1-x)}\,dx.$

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- Apr 12th 2008, 05:22 PMKrizalidToday's calculation of integral #9
Find $\displaystyle \int_0^1\frac{x^2}{(1-x)\ln^2(1-x)}\,dx.$

- Apr 12th 2008, 08:24 PMMathstud28Ok
You can rewrite it as $\displaystyle \int_0^1{\sum_{n=0}^{\infty}x^{n+2}}\cdot\bigg(-\sum_{n=0}^{\infty}\frac{x^{n}}{n}\bigg)^{-2}$...which then is equal to$\displaystyle \int_0^1{\sum_{n=0}^{\infty}x^{n+2}\cdot{2\sum_{n= 0}^{\infty}\sum_{n=0}^{\infty}\frac{n^2}{x^{2n}}}}$...and then $\displaystyle 2\int_0^1{\sum_{n=0}^{\infty}\sum_{n=0}^{\infty}\s um_{n=0}^{\infty}\frac{x^{2-n}}{n}}$...so wed get $\displaystyle 2\sum_{n=0}^{\infty}\sum_{n=0}^{\infty}\sum_{n=0}^ {\infty}\frac{x^{3-n}}{n(3-n)}$ evaluated from 0 to 1...I know I know I did it completely wrong...but will someone point out where? I just did this for novelty (Rofl)

- Apr 13th 2008, 05:20 AMKrizalid
- Apr 13th 2008, 08:01 AMThePerfectHacker
I did not solve it yet. Here is the first step I was thinking. Let $\displaystyle t = -\ln (1-x)$.

This gives us, $\displaystyle \int_0^{\infty} \left( \frac{1 - e^{-t}}{t} \right)^2 dt$. - Apr 13th 2008, 08:04 AMMoo
Hello,

Let $\displaystyle u=\ln(1-x)$

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$\displaystyle du=-\frac{dx}{1-x}$

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$\displaystyle e^u=1-x$

$\displaystyle x=1-e^u$

$\displaystyle x^2=(1-e^u)^2$

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If x=0 then $\displaystyle u=0$

If x=1 then $\displaystyle u=-\infty$

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Inverting the extremities of the integral with the - sign coming from du

$\displaystyle \int_0^1\frac{x^2}{(1-x)\ln^2(1-x)} dx = \int_{-\infty}^0 \frac{(1-e^u)^2}{u^2} du$

Where did i go wrong ? :'( - Apr 13th 2008, 08:08 AMThePerfectHacker
- Apr 13th 2008, 08:11 AMMoo
Actually, i've been trying but gave up an hour ago... The first parts of integration by parts were divergent oO

And ok, when i posted i saw the first version of your integral and got anxious because it was not the same... - Apr 13th 2008, 08:23 AMMathstud28So is it this?
is that then equal to $\displaystyle \int{\bigg(-2\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n-1}}{n!}+\sum_{n=0}^{\infty}\frac{(-2)^{n}x^{n-2}}{n!}+1\bigg)dx}$$\displaystyle =-2\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n}}{n\cdot{n!}}+\sum_{n=0}^{\infty}\frac{ (-2)^{n}x^{n-1}}{(n-1)n!}+x$ evaluated from 0 to ∞?

- Apr 13th 2008, 08:33 AMMoo
Yop,

Where did $\displaystyle \frac{1}{t^2}$ go ? :p (it's it, instead of +1)

But your method doesn't simplify the problem... or i've missed a point ^^ - Apr 13th 2008, 08:38 AMMathstud28oops
- Apr 13th 2008, 10:26 AMKrizalid
- Apr 16th 2008, 06:22 PMKrizalid
Hint: construct a double integral. (Following on post #4.)

- Apr 26th 2008, 12:06 PMKrizalid
What we need is $\displaystyle \frac{1}{t^{2}}=\int_{0}^{\infty }{xe^{-tx}\,dx},\,t>0.$ From there you may observe that the conclusion follows.

- Apr 27th 2008, 04:45 PMThePerfectHacker
- Apr 29th 2008, 08:39 AMThePerfectHacker
In general, $\displaystyle \frac{(n-1)!}{t^n} = \int_0^{\infty} x^{n-1} e^{-tx} dx$ for $\displaystyle t>0$ and $\displaystyle n\geq 1$. This seems to be a useful fact to know in double integration.